Find $a_{2012}-3a_{2010}/3 a_{2011}$ where the sequence $a_n$ is determined by roots of a quadratic equation

Question

If $\alpha$ and $\beta$ are the roots of $x^2-9x-3=0$, $a_n=\alpha^n-\beta^n$ and $b_n=\alpha^n+\beta^n$, then find the value of $\dfrac{a_{2012}-3a_{2010}}{3 a_{2011}}$ and $\dfrac{b_{2012}-3b_{2010}}{3b_{2011}}$.

Please help me solve this problem. I tried finding relation between a1, a2, a3 and b1, b2, b3. But I could go no further.

Answer 1

We have$$a_{2012}-3a_{2010}=(\alpha^{2012}-\beta^{2012})-3(\alpha^{2010}-\beta^{2010})=$$

$$=\alpha^{2010}(\alpha^2-3)-\beta^{2010}(\beta^2-3)$$

But $\alpha^2-3=9\alpha$ and $\beta^2-3=9\beta$, since each of them satisfy the quadratic equation. Hence we may simplify as $$\alpha^{2010}(9\alpha) - \beta^{2010}(9\beta)=9(\alpha^{2011}-\beta^{2011})=9a_{2011}$$

Dividing by $3a_{2011}$ gives just 3. The other problem is similar.

Answer 2

$\alpha^2=9\alpha+3$ implies $\alpha^{n+2}=9\alpha^{n+1}+3\alpha^n$. Ditto for $\beta$. Hence $a_{n+2}=9a_{n+1}+3a_n$. Ditto for $b_n$.